As you must said, (0, 0) makes both numerator and denominator 0 so (0, 0) is the only critical point.
I personally find it simpler to think of problems like these as two separate equations. If
\frac{dy}{dx}= \frac{F(x, y)}{G(x,y)}
then, introducing the parameter t, we must have
\frac{dy}{dt}= F(x,y)
\frac{dx}{dt}= G(x,y)
If we think of t as "time", we can think of these equations as describing motion in the xy plane.
"Critcal points" (or, as I would say, "equilibrium" points) are where everything is "balanced"- neither x nor y changes: F(x,y)= 0, G(x,y)= 0.
Now, what happens if x and y are close to a critical point? dx/dt and dy/dt are not 0 so the point will "move" but will it move toward or away from the critical point? If such a point always moves closer to the critical point, that is a stable point. If it can move away from the critical point, that is an unstable point.
Here, you critical point is (0, 0) and your "equations of motion" are
\frac{dx}{dt}= 4xy
and
\frac{dy}{dt}= -(2x+ y^2)
If x is a small positive number and y is 0, that gives
\frac{dx}{dt}= 0
\frac{dy}{dt}= -2x
so the "motion" is really a circulation, not going toward or away from (0, 0).
If y is a small positive number and x is 0, we have
\frac{dx}{dt}= 0
\frac{dy}{dt}= -y^2
downward, so now the point is moving toward (0, 0).
But if y is a small negative number and is 0, because of that square, we have
\frac{dx}{dt}= 0
\frac{dy}{dt}= -y^2
still downward but now way from (0, 0).
Because it is possible to move away, this is an unstable point.
